An object 4cm high is placed on the principal axis with a distance of 20cm in front of a convex lens of focal length 30cm. Find the distance of image from the lens its height and its nature.

Solution:
Given,Object distance (u) = -20 cm focal length (f) = 30 cm image distance (v) = ?Formula used :1f= 1v - 1u1v = 1f + 1u1v = 130 + 1(-20)  = 130- 120 = -160v = -60 cmMagnification = vu = -60-20= 3Image would be virtual, enlarged and erect.Height of image will behiho =3hi = 3 ho = 3×4 = 12 cm

 

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